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\begin{center}
\large\bf
36-303: Sampling, Surveys and Society\\
Exam 2 Solutions ---
Tue Apr 13, 2010
\end{center}

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\def\ans{\hspace*{-1.8em}\fbox{\phantom{iii.}}}

\begin{enumerate}

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\item\pts{20} {\em Multiple Choice (4 parts).  }

\begin{enumerate}

\item\pts{5} {\em The {\em bias} between $\bar Y_R$ and $\bar
  Y_{pop}$, due to missing responses, is
\[
    \bar Y_R - \bar Y_{pop} = \frac{N_M}{N} (\bar Y_R - \bar Y_M)
\]
Which statement below is false (or, circle iv. if all are OK)?}
\begin{enumerate}\small
\item\ans All other things being equal, the more people who
  would {\em not respond}, the smaller the bias due to missing
  responses.
\item All other things being equal, the bigger the difference between
  mean number of meetings attended by nonresponders, vs the mean
  number attended by responders, the bigger the bias due to
  missing responses.
\item The larger your SRS, the better you can estimate $\bar Y_R$.
\item All of the above statements are true.
\end{enumerate}

\item\pts{5} {\em In one-stage clustered sampling, the ICC $\rho$ measures}
\begin{enumerate}\small
\item The correlation between observations in different clusters.
\item The correlation between the cluster means of different clusters.
\item \ans The correlation between observations in the same cluster.
\item The correlation between the cluster mean and the individual
  observations in the cluster.
\end{enumerate}

\item\pts{5} {\em Some respondents are never presented
  with the last page of the survey, due to a bug in the website
  software.  This is most likely an example of}
\begin{enumerate}\small
\item Unit nonresponse, Missing At Random (MAR)
\item \ans Item nonresponse, Missing Completely At Random (MCAR)
\item Unit nonresponse, Missing Completely At Random (MCAR)
\item Item nonresponse, Missing At Random (MAR)
\end{enumerate}

\item\pts{5} {\em If {\em observations {\bf within groups} are more
    {\bf similar} to each other}, and {\em observations {\bf between
      groups} are more {\bf different} from each other}, then, all
  other things being equal, we expect}
\begin{enumerate}\small
\item The variance of the stratified sample mean $\bar y_{st}$ will go
  {\bf up} and the variance of the cluster sample mean $\bar y_{cl}$
  will go {\bf down}.
\item \ans The variance of the stratified sample mean $\bar y_{st}$ will go
  {\bf down} and the variance of the cluster sample mean $\bar y_{cl}$
  will go {\bf up}.
\item Both variances will go {\bf up}.
\item Both variances will go {\bf down}.
\end{enumerate}

\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\item\pts{20} {\em Stratified Sampling (3 parts).}
% From Dick Scheaffer book, pp 61 ff

\begin{center}
\begin{tabular}{rccc}
\hline
                       & \em Stratum 1 & \em Stratum 2 & \em Stratum 3 \\
\hfill                 & \em Alpha     & \em Berea     & \em Rural Area\\
\hline
\hline
\hfill \em Number of households  & $N_1=155$ & $N_2=62$  & $N_3=93$\\
\hfill \em Stratum sample size   & $n_1=20$  & $n_2=8$   & $n_3=12$\\
\hline
\em TV viewing time, hr/wk & 
\begin{tabular}[t]{r@{~~}r@{~~}r@{~~}r}
35 & 28 & 26 & 41 \\
43 & 29 & 32 & 37  \\
36 & 25 & 29 & 31  \\
39 & 38 & 40 & 45 \\
28 & 27 & 35 & 34 
\end{tabular}
&
\begin{tabular}[t]{r@{~~}r@{~~}r@{~~}r}
27 &  4 & 49 & 10 \\
25 & 41 & 25 & 30 
\end{tabular}
&
\begin{tabular}[t]{r@{~~}r@{~~}r@{~~}r}
 8 & 15 & 21 &  7 \\
14 & 30 & 20 & 11 \\
12 & 32 & 34 & 24 
\end{tabular}
\\
\hline
\em Stratum Statistics & $\bar y_1 = 33.9000$ & $\bar y_2 = 25.125$ &
$\bar y_3 = 19.000$ \\
                   & $s^2_1 = 35.358$ & $s^2_2 = 232.411$ & $s^2_3 = 87.636$\\
\hline
\end{tabular}
\end{center}

\hfil \fbox{\parbox{5in}{{\em Note:} The given value of $\bar y_2$ is
    incorrect, given the data in the table (the correct value was
    26.375).  However, I asked everyone to use the given value.}}

\begin{enumerate}


\item\pts{6} {\em  Compute $\bar y_{st}$ and $\bar y_{srs}$.  If they
  should be the same, explain why.  If they should be different,
  explain why.}

We first calculate the stratum weights,
\[
	W_1 = N_1/N = 155/310 = 0.5 ;~~ W_2 = N_2/N = 62/310 = 0.2 ;~~ W_3 =
        N_3/N = 93/310 =0.3 .
\]
Then we calculate
\[
	\bar y_{st} = W_1\bar y_1 + W_2\bar y_2 + W_3\bar y_3 
        = (0.5)(33.9) + (0.2)(25.125) + (0.3)(19.000) = \fbox{27.675}
\]
\fbox{The value for $\bar y_{srs}$ is the same}, because this design
is self-weighted\footnote{A stratified design with the same sampling
  fraction $f$ in each stratum is self-weighted.  See answer to part
  (b) also.}.  You can verify that by calculating $\bar y_{srs}$
explicitly.



\item\pts{6} {\em Compute SE$(\bar y_{st})$ and use this to create an
  approximate 95\% confidence interval for the mean household TV
  viewing time (hrs/wk) in Jefferson County.}

We need the sampling fractions
\[
    f_1 = n_1/N_1 = 20/155 = 0.129; ~~ f_2 = n_2/N_2 = 8/62 = 0.129;
    ~~ f_3 = n_3/N_3 = 12/93 = 0.129
\]
The variance is then
\begin{eqnarray*}
SE(\bar y_{st})^2 & = & W_1 (1-f_1) s^2_1/n_1 + W_2 (1-f_2) s^2_2/n_2 +
   W_3 (1-f_3) s^2_3/n_3 \\  
   &= & (0.5)(1-0.129)(35.358/20) + (0.3)(1-0.129)(232.411/8) \\&& ~+~
   (0.2)(1-0.129)(87.636/12) ~ = ~ 9.633 
\end{eqnarray*}
and the standard error is
\[
	SE(\bar y_{st}) = \sqrt{9.633} = \fbox{3.104},
\]
so the 95\% CI is about $27.675 \pm 2\times 3.104$, or \fbox{$(21.467,
  33.883 )$}.



\item\pts{8} {\em If we were to treat this as an SRS rather than a
  stratified sample, the sample variance would be $s^2=83.658$. 
\begin{enumerate}
\item Estimate the DEFF (design effect) for this design.
\item Was it a good idea to do the stratified sample rather than an
  SRS?  Why or why not?
\end{enumerate}
}

The DEFF is 
\[
     DEFF ~ = ~ \frac{\Var(\overline y_{st})}{\Var(\overline y_{srs})}
	= \frac{9.633}{83.658} = \fbox{0.12}
\]
Since $\Var(\overline y_{st})$ is almost 10 times smaller than
$\Var(\overline y_{srs})$, it is \\
\fbox{definitely better to do the
  stratified design.}

\hfil\fbox{\parbox{5in}{{\em Note:} $Var(\bar y_{srs})$ is actually
    wrong here.  It should have been $Var(\bar y_{srs}) = 11.136$,
    instead of 83.658.  In
    that case, DEFF = 9.633/11.136 = 0.87.  This is still a win for
    the stratified sampling design.}} 

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item\pts{18} {\em Approval Rating of Sen. Diane Feinstein, Part I (3 parts).}

{\em Approximately equal numbers of Democrats (342) and Republicans (298)
were contacted by phone; 267 of the Democrats and 98 of the
Republicans approved of the job Sen. Feinstein was doing.

According to the Ballot Access News
(http://www.ballot-access.org/2008/120108.html), there were 7,683,495
registered Democrats and 5,428,052 registered Republicans in
California at approximately the same time as the Field Poll was being
taken.
	
}


\begin{enumerate}

\item\pts{6} {\em Compute $\hat p_{dem}$, the proportion of Democrats
  who approve of Sen. Feinstein, and its standard error, ${\rm
    SE}(\hat p_{dem})$.}

\begin{eqnarray*}
	\hat p_{dem} & = & 267/342 = \fbox{0.78}\\
        SE(\hat p_{dem}) & = & \sqrt{(1-f)\hat p_{dem}(1-\hat
          p_{dem})/(n_{dem}-1)} \\
          & = & \sqrt{(1-342/7683495)(0.78)(1-0.78)/(342-1)}\\
          & = & \fbox{0.022}
\end{eqnarray*}

\item\pts{6} {\em Do the same for Republicans: compute $\hat p_{rep}$ and
  ${\rm SE}(\hat p_{rep})$.}

\begin{eqnarray*}
	\hat p_{rep} & = & 98/298 = \fbox{0.33}\\
        SE(\hat p_{rep}) & = & \sqrt{(1-f)\hat p_{rep}(1-\hat
          p_{rep})/(n_{rep}-1)} \\
          & = & \sqrt{(1-298/5428052)(0.33)(1-0.33)/(298-1)}\\
          & = & \fbox{0.027}
\end{eqnarray*}

\item\pts{6} {\em Using the SE for the difference, $\sqrt{{\rm
      SE}(\hat p_{dem})^2 + {\rm SE}(\hat p_{rep})^2}$, compute an
  approximate 95\% CI for the difference between the true proportions
  of Democrats and Republicans who approve of Senator Feinstein.}

The combined SE is
\[
	\sqrt{{\rm SE}(\hat p_{dem})^2 +
  {\rm SE}(\hat p_{rep})^2} = \sqrt{(0.022)^2+(0.027)^2} = 0.035 
\]
and the difference vetween the two $\hat p$'s is 
\[
    \hat p_{dem} - \hat p_{rep} = 0.78 - 0.33 = 0.45
\]
so the CI is $0.45 \pm 2\times 0.035$, or \fbox{(0.38,0.52)}.

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item\pts{24} {\em Approval Rating of Sen. Diane Feinstein, Part II (4
  parts).}

\begin{enumerate}

\item\pts{6} {\em Combining Democrats and Republicans, estimate the overall
  proportion of voters in California who approve.}

$\hat p = ( 267 + 98)/( 342+298) = \fbox{0.57}$

\item\pts{6} {\em Using political party registration to define two
  post-strata, compute post stratification weights to apply to this
  survey.}

Note that the total population size is $N=N_{dem}+N_{rep} =
13,111,547$, and the total sample size is $n=n_{dem} + n_{rep} = 640$.

Using this information, the weights are then
\begin{eqnarray*}
  \mbox{\rm For Democ. $i$'s:}~~
  w_i & = & (N_{dem}/N)/(n_{dem}/n) = (7,683,495/13,111,547)/(342/640) = \fbox{1.097}\\
  \mbox{\rm For Repub. $i$'s:}~~
  w_i & = & (N_{rep}/N)/(n_{rep}/n) = (5,428,052/13,111,547)/(298/640) = \fbox{0.889}
\end{eqnarray*}

\item\pts{6} {\em Compute the
  weighted proportion of voters in California who approve of
  Sen. Feinstein's job.} 
\[
    \overline y_w = \frac{\sum_i w_i y_i}{\sum_i w_i}  =
    \frac{(1.097)(267\times1 + 75\times0) + (0.889)(98\times1+200\times0)}{(1.097)(342) + (0.889)(298)}
    = \fbox{0.59}
\]
since each $y_i$ is either 1 (approve) or 0 (disapprove).

\item\pts{6} {\em Use the values of the weights to explain the difference
  between your answers in parts (a) and (c).}

The weight for Democratic respondents, 1.097, is larger than the
weight for Republican respondents, 0.889 (to make the Democratic and
Republican contributions to $\bar y_w$ proportional to their
population counts).  Since 1.097$>$0.889, the favorable Democratic
responses will count a bit more than the disfavorable Republican
responses, leading to a higher estimated proportion of voters
approving of Sen. Feinstein.


\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Examples from pp 167-168 of blue Groves book...

% Examples from lecture 09...

\item\pts{18} {\em Data Collection Methods (3 parts.)} {\em In each
  situation below, choose the best mode of data collection, and give a
  reason why.}

\hfil\fbox{\parbox{5in}{Note: Below I have given the intended mode of
    data collection and ``model'' reasons for those answers.  You
    might come up with other reasons for these answers.  You might
    even come up with different modes of data collection.  Your answer
    and/or reason will be counted correct, if different from mine,
    {\bf as long as} you provide a {\bf clear and convincing reason}.
    The grader's decisions (or mine, if he asks for my help) will be
    final.}}

\begin{enumerate}

\item\pts{6} {\em Your market research firm has been asked to conduct a
  survey in Pittsburgh to see who would purchase a new, sweet, fruity
  carbonated soft drink.}
\begin{itemize}
\item {\em Which mode of data collection (circle the best answer)?}
\begin{enumerate}
\item Telephone survey.
\item \ans Face to Face interviews.
\item Internet (web-based) survey.
\end{enumerate}
\item {\em Why?}

Here are some possible considerations:
\begin{itemize}
\item An Internet survey would not reach enough of the target population
(almost anyone buys sodas, but not everyone has consistent internet
access).  This could introduce some bias into the survey, since we
would not have opinions of people who for economic or other reasons
did not have internet access.
\item A telephone survey would reach most households and consumers.
  This is a good, not too expensive choice.  A drawback of this choice
  is that you can't have consumers actually try to the new soda; and
  this is information consumers might need before deciding whether
  they would purchase it.
\item Face to face interviews are expensive but you can have consumers
  try the product directly, rather than asking them  to respond to a
  completely hypothetical question.
\end{itemize}


\end{itemize}

\item\pts{6} {\em The Governor of your state has asked you to design a
  survey to assess the views of residents of the state toward public
  transportation and to evaluate how likely they would be to use
  alternative modes of transportation.}
\begin{itemize}
\item {\em Which mode of data collection (circle the best answer)?}
\begin{enumerate}
\item \ans Telephone survey.
\item Face to Face interviews.
\item Internet (web-based) survey.
\end{enumerate}
\item {\em Why?}

Here are some possible considerations:
\begin{itemize}
\item An Internet survey would not reach enough of the target
  population (almost anyone has to consider modes of transportation,
  but not everyone has consistent internet access).  This could
  introduce some bias into the survey, since we would not have
  opinions of people who for economic or other reasons might not have
  internet access (and this might be correlated with transportation
  choices).
\item Face to face interviews are expensive and there is almost
  no information on this topic that you could provide the respondent
  with in a face to face setting, that you could not also provide on
  the phone. 
\item A telephone survey would reach most people in the state (good
  coverage of the target population by the sampling frame).  This is a
  good, not too expensive choice.
\end{itemize}

\end{itemize}


\item\pts{6} {\em The ASA wishes to do a survey of its members across
  the United States on their views toward certification of
  statisticians (statisticians would need pass a test in order to get
  a license to practice statistics, like doctors, some engineering
  professions, etc.).  }
\begin{itemize}
\item {\em Which mode of data collection (circle the best answer)?}
\begin{enumerate}
\item Telephone survey.
\item Face to Face interviews.
\item \ans Internet (web-based) survey.
\end{enumerate}
\item {\em Why?}

Here are some possible considerations:
\begin{itemize}
\item Face to face interviews would be prohibitively expensive for a
  nationwide survey.
\item A telephone survey would reach most members of the ASA.  This is
  a good choice, but\ldots
\item An Internet survey would be much cheaper than a face-to-face
  survey, and also cheaper than a telephone survey.  If there is
  a great deal of text to consider (e.g. a description of the
  certification process) it is better presented on a web page than
  verbally over the phone.  Moreover, since most statisticians are
  white collar workers with ready access to the internet, few
  statisticians would be unable to complete the survey for lack of
  access to the survey website.
\end{itemize}

\end{itemize}

\end{enumerate}


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\comment{

\clearpage

% Examples from lecture 09...

\item\pts{??} Another 36-303 group wants to do a survey of PAT bus
  riders' attitudes toward a proposed schedule of bus fare increases.
  In their survey proposal they specify the {\em target population} to
  be all adult residents of Pittsburgh's East End (Oakland, Shadyside,
  Squirrell Hill \& Point Breeze).  They plan to collect data by
  approaching people waiting at six specific bus stops near where
  these students live, between 8am and 10am for one whole work-week
  (Mon-Fri), for face-to-face interviews.
\begin{enumerate}

\item\pts{??} Did this group specify a {\em sampling frame} for their
  survey?  If so, say what it is.  If not, specify a sampling frame
  that the group could work with.

\vspace*{\fill}

\item\pts{??} Write down two very likely sources of {\em coverage
  error} for this survey as currently proposed.

\vspace*{\fill}

\item\pts{??} Make one big suggestion that this group could implement
  to improve their sampling plan.

\vspace*{\fill}

\end{enumerate}

}

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\end{enumerate}

\end{document}

